Acceleration due to gravity on Earth is approximately?
Show answer & explanation
Correct answer: B
Explanation
Solution: The correct answer is 9.8 m/s².
Explanation:
☛ Acceleration due to gravity (g) is the acceleration gained by an object due to the gravitational force exerted by the Earth.
☛ On the surface of the Earth, the standard average value of acceleration due to gravity is approximately 9.8 m/s² (or 9.80665 m/s²).
☛ The value of 'g' is calculated using the formula g = GM / R², where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
☛ It is a vector quantity, and its direction is always towards the centre of the Earth.
Extra Exam-Relevant Facts:
☛ Variation at Poles and Equator: The value of 'g' is maximum at the poles (approx. 9.83 m/s²) and minimum at the equator (approx. 9.78 m/s²) due to the equatorial bulge and Earth's rotation.
☛ Altitude and Depth: The value of 'g' decreases both with an increase in height (altitude) above the Earth's surface and with an increase in depth below the surface.
☛ Centre of the Earth: The value of acceleration due to gravity at the centre of the Earth is zero.
☛ Universal Gravitational Constant (G): Its value is 6.674 × 10⁻¹¹ N·m²/kg², which remains constant throughout the universe.
Important One-Liners for Upcoming Exams:
☛ What is the SI unit of acceleration due to gravity? → m/s² (metre per second squared)
☛ What is the approximate value of acceleration due to gravity on the Moon? → 1/6th of that of Earth (approx. 1.62 m/s²)
☛ What is the weight of an object at the centre of the Earth? → Zero
☛ Who accurately measured the Universal Gravitational Constant (G) for the first time? → Henry Cavendish (1798)
☛ How does the value of 'g' change if the Earth suddenly stops rotating? → The value of 'g' will increase at the equator (remains unchanged at poles)